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Then considering for all
x
∈
[
−
N, N
], one can write
f
(
x
)as
N−
1
c
j
j
)
n−
1
+
f
(
x
)=
f
[
−N,−N
+1)
(
x
)+
1)!
(
x
−
,
(8.3)
(
n
−
j
=
−N
+1
where we use the following notation,
x
+
=
max
(0
,x
)
,
(8.4)
x
n−
1
+
=(
x
+
)
n−
1
,
n
≥
2
.
The collection of
n
+2
N
−
1 functions
1)
n−
1
+
N
+1)
n−
1
+
x
0
,x
1
,
,x
n−
1
,
(
x
+
N
{
···
−
,
···
,x
−
}
(8.5)
,x
n−
1
by truncated
is a basis of
S
n,N
. We can replace the monomials 1
,x,
···
powers:
1)
n−
1
+
,
(
x
+
N
)
n−
1
+
(
x
+
N
+
n
−
,
···
.
With this, we now can generate the entire set of truncated powers by integer
translates of a single function,
x
n−
1
+
as,
r
)
n−
1
+
{
(
x
−
,r
=
−
N
−
n
+1
,
···
,N
−
1
}
.
(8.6)
This is also a basis of
S
n,N
.
Now for different values of
N
, different spaces
S
n,N
can be visualized; each
of them is of finite dimension when
N
is finite. Making
N
infinitely large and
considering the union of all such spaces, we can make the space
S
n
of infinite
dimension and the basis in equation (8.6) will, therefore, be a different basis
of the infinite dimensional space
S
n
(due to the different bases for the values
of
N
). This basis can be written as
r
)
n−
1
+
B
1
=
{
(
x
−
,
r
∈
Z.
(8.7)
To find cardinal splines in
L
2
(
IR
), one can consider backward differences
with recursion as
(
f
)(
x
)=
f
(
x
)
−
f
(
x
−
1)
(8.8)
k
f
)(
x
)=(
k
−
1
(
(
f
))
,
k
=2
,
3
,
···
For
n
th order polynomial, the
n
th order difference is zero, i.e.,
n
f
=0
,
f
∈
π
n−
1
.
Let us now define a linear combination of the basis functions given in equation
(8.7) as
1
n
x
n−
1
+
M
n
(
x
)=
1)!
,
n
≥
2
(8.9)
(
n
−
where
M
1
=
N
1
, the characteristic function of [0
,
1), i.e.,
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